$2^\circ$ Suppose that there is a choice function $\Zobr fn{\bigcup A_i}$. We want to extend it to a new function $g$ defined on $n+1=n\cup\{n\}$ in a such way that $g(n)\in A_n$. Again we know that there exists $a\in A_n$ and we put $g=f\cup\{(n,a)\}$. (Again, this is application of existential instantiation.)